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Chapter 6: Problem 79

A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and then one of the pair is assigned to each of the two treatments. The treatments are applied, and each is either a success (S) or a failure (F). The researchers keep track of the total number of successes for each treatment. They plan to continue the chance experiment until the number of successes for one treatment exceeds the number of successes for the other treatment by \(2 .\) For example, they might observe the results in the table below. The chance experiment would stop after the sixth pair, because Treatment 1 has 2 more successes than Treatment \(2 .\) The researchers would conclude that Treatment 1 is preferable to Treatment \(2 .\) Suppose that Treatment 1 has a success rate of \(.7\) (i.e., \(P(\) success \()=.7\) for Treatment 1 ) and that Treatment 2 has a success rate of \(.4\). Use simulation to estimate the probabilities in Parts (a) and (b). (Hint: Use a pair of random digits to simulate one pair of subjects. Let the first digit represent Treatment 1 and use \(1-7\) as an indication of a

Short Answer

Expert verified
It's not possible to provide a numerical short answer for this question because it depends on the simulation outcome. It must be noted that the expected probability for Treatment 1 is higher than Treatment 2. This might lean towards a conclusion where Treatment 1 stops first, with 2 more successes than Treatment 2. The exact numerical probabilities would depend on the simulated experiments.

Step by step solution

01

Understanding the problem

The purpose of this exercise is to determine probabilities of success for two different treatments, Treatment 1 and Treatment 2. Their success rates are given as .7 and .4 respectively. The result interpretation is to be made when the number of successes of one treatment exceeds the other's by 2.
02

Simulating results for Treatment 1

Let's simulate a large number of experiments (say, 10,000) for Treatment 1 using random numbers. The digits between 1 and 7 will be considered a 'success', simulating the .7 success rate or 70% probability of Treatment 1. Count the number of successes.
03

Simulating results for Treatment 2

Execute the same process for Treatment 2. This time, consider the digits between 1 and 4 as 'success', which would simulate the .4 success rate or 40% probability of Treatment 2. Keep track of the number of successes.
04

Comparison and Concluding Results

Compare the amount of successes of both treatments. The experiment ends when the number of successes for one treatment exceed the other by 2. Analyze the simulation results and compute the probabilities. This can be given by the count of experiments where one treatment exceeds the other by 2 successes over the total count of simulated experiments.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Estimation
Probability estimation is an essential concept in statistical simulation, often used to predict outcomes in uncertain scenarios. In this exercise, the research team aims to estimate the likelihood that one treatment will have 2 more successes than the other. They achieve this by employing random digit simulations to predict outcomes.
Utilizing random numbers helps mimic the real-world randomness, making estimations more accurate and reliable. In this context:
  • For Treatment 1, a random digit from 1 to 7 represents a successful outcome, accounting for its 70% success rate.
  • For Treatment 2, a successful outcome is represented by a random digit from 1 to 4, aligning with its 40% success rate.
Simulating thousands of such instances provides a robust estimate of the probability that one treatment will outperform the other by a definitive margin. This approach ensures reliable predictions that are invaluable for medical researchers.
Medical Research
Medical research frequently relies on statistical methods to evaluate treatment efficiency and patient outcomes. This exercise exemplifies how a structured approach can guide researchers in making informed decisions.
Simulations are immensely beneficial in scenarios where actual trials might be costly, time-consuming, or ethically challenging. By creating a model of the situation using simulation, researchers can efficiently:
  • Assess potential treatment outcomes without real-world trials.
  • Understand variables that might affect the success rates.
  • Predict trends based on statistical probability, aiding in treatment decisions.
Medical research thus strongly benefits from statistical simulation by providing a hypothetical framework that approximates real-life scenarios, allowing for a deeper understanding of the potential effectiveness of treatments.
Success Rate
Success rate is a critical metric in evaluating any medical treatment. It provides a numerical measure of how effective a treatment is in achieving desired outcomes. In this scenario:
  • Treatment 1 demonstrates a higher success rate of 70%, indicating a higher probability of a positive outcome for patients receiving this treatment.
  • Treatment 2, with a success rate of 40%, indicates a relatively lower chance of achieving the desired therapeutic effect.
Such metrics are crucial in determining which treatments are more effective and under what conditions. It helps researchers and medical professionals decide which treatment to adopt or recommend for patient care, aiming to maximize positive outcomes and efficiency.
Comparison of Treatments
Comparing treatments is integral in determining the most effective strategy for medical interventions. The exercise provides a framework for evaluating two treatments based on success rates and observed outcomes through simulation.
Assessing the differences in outcome helps:
  • Identify which treatment is likely to be more successful under certain conditions.
  • Guide resource allocation by determining where efforts should be focused to improve patient care.
  • Facilitate the development of new, more effective treatments based on observed data trends between different options.
Conducting such comparisons through simulation is a cost-effective method of gaining insight into treatment performance prior to real-world application, offering researchers a clear picture of potential benefits and drawbacks.

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Most popular questions from this chapter

A family consisting of three people- \(\mathrm{P}_{1}, \mathrm{P}_{2}\), and \(\mathrm{P}_{3}\) \- belongs to a medical clinic that always has a physician at each of stations 1,2, and 3 . During a certain week, each member of the family visits the clinic exactly once and is randomly assigned to a station. One experimental outcome is \((1,2,1)\), which means that \(\mathrm{P}_{1}\) is assigned to station \(1 .\) \(\mathrm{P}_{2}\) to station 2, and \(\mathrm{P}_{3}\) to station 1 a. List the 27 possible outcomes. (Hint: First list the nine outcomes in which \(\mathrm{P}_{1}\) goes to station 1 , then the nine in which \(\mathrm{P}_{1}\) goes to station 2 , and finally the nine in which \(\mathrm{P}_{1}\) goes to station 3 ; a tree diagram might help.) b. List all outcomes in the event \(A\), that all three people go to the same station. c. List all outcomes in the event \(B\), that all three people go to different stations. d. List all outcomes in the event \(C\), that no one goes to station \(2 .\) e. Identify outcomes in each of the following events: \(B^{C}\), \(C^{C}, A \cup B, A \cap B, A \cap C\)

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7, P(F)=.6\), and \(P(E \cap F)=.54\). a. Calculate \(P(E \mid F)\) the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

An article in the New York Times (March 2, 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P\) (survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

The Associated Press (San Luis Obispo TelegramTribune, August 23,1995 ) reported on the results of mass screening of schoolchildren for tuberculosis (TB). For Santa Clara County, California, the proportion of all tested kindergartners who were found to have TB was .0006. The corresponding proportion for recent immigrants (thought to be a high-risk group) was .0075. Suppose that a Santa Clara County kindergartner is selected at random. Are the events selected student is a recent immigrant and selected student has \(T B\) independent or dependent events? Justify your answer using the given information.

The Los Angeles Times (June 14,1995 ) reported that the U.S. Postal Service is getting speedier, with higher overnight on-time delivery rates than in the past. The Price Waterhouse accounting firm conducted an independent audit by seeding the mail with letters and recording ontime delivery rates for these letters. Suppose that the results were as follows (these numhers are fictitions hut are compatible with summary values given in the article): $$ \begin{array}{lcc} & \begin{array}{l} \text { Number } \\ \text { of Letters } \\ \text { Mailed } \end{array} & \begin{array}{l} \text { Number of } \\ \text { Lefters Arriving } \\ \text { on Time } \end{array} \\ \hline \text { Los Angeles } & 500 & 425 \\ \text { New York } & 500 & 415 \\ \text { Washington, D.C. } & 500 & 405 \\ \text { Nationwide } & 6000 & 5220 \\ & & \\ \hline \end{array} $$ Use the given information to estimate the following probabilities: a. The probability of an on-time delivery in Los Angeles b. The probability of late delivery in Washington, D.C. c. The probability that two letters mailed in New York are both delivered on time d. The probability of on-time delivery nationwide
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